# In an isosceles triangle, the height drawn to the base is 16 and the radius of the inscribed circle is 6

**In an isosceles triangle, the height drawn to the base is 16 and the radius of the inscribed circle is 6, find the perimeter of the triangle.**

Let’s draw the radius of the circle OK, which is perpendicular to the side AB, then the triangle OBK is rectangular.

The length of the height AH = 16 cm, and OH = 6 cm as the radius of the circle, then OB = 16 – 6 = 10 cm.

From the triangle BOK BK ^ 2 = OВ ^ 2 – OK ^ 2 = 100 – 36 = 64.

BK = 8 cm.

AK = AH as tangents to the circle drawn from one point. Let AK = AH = X cm.

Then AB = X + 8 cm.

From a right-angled triangle ABH, AB ^ 2 = BH ^ 2 + AH ^ 2.

(X + 8) ^ 2 = 256 + X ^ 2.

X ^ 2 + 16 * X + 64 = 256 + 2.

16 * X = 192.

X = AH = AK = 192/16 = 12 cm.

Then AB = BC = 8 + 12 = 20 cm.

AC = 2 * AH = 2 * 12 = 24 cm.

Determine the perimeter of the triangle. Ravs = AB + BC + AC = 20 + 20 + 24 = 64 cm.

Answer: The perimeter of the triangle is 64 cm.